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A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of the photon.

A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 level. Estimate the frequency of the photon

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#### Solution

For ground level, n_{1} = 1

Let E_{1} be the energy of this level. It is known that E_{1} is related with n_{1} as:

`"E"_1 = (-13.6)/("n"_1^2) "eV"`

`= (-13.6)/1^2`

= −13.6 eV

The atom is excited to a higher level, n_{2} = 4.

Let E_{2} be the energy of this level.

∴ `"E"_2 = (-13.6)/"n"_2^2 "eV"`

= `(-13.6)/4^2`

= `(-13.6)/16 "eV"`

The amount of energy absorbed by the photon is given as:

E = E_{2} − E_{1}

= `(-13.6)/16 - ((-13.6)/1)`

= `(13.6 xx 15)/16 "eV"`

= `(13.6 xx 15)/16 xx 1.6 xx 10^(-19)`

= 2.04 × 10^{−18} J

For a photon of wavelength λ, the expression of energy is written as:

`"E" = "hc"/lambda`

Where,

h = Planck’s constant = 6.6 × 10^{−34} Js

c = Speed of light = 3 × 10^{8} m/s

∴ λ = `"hc"/"E"`

= `(6.6 xx 10^-34 xx 3 xx 10^8)/(2.04 xx 10^-18)`

= 9.7 × 10^{−8} m

= 97 nm

And, frequency of a photon is given by the relation,

`"v" = "c"/lambda`

= `(3 xx 10^8)/(9.7 xx 10^(-8))`

≈ 3.1 × 10^{15} Hz

Hence, the wavelength of the photon is 97 nm while the frequency is 3.1 × 10^{15} Hz.

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